Answer
$\dfrac{65}{28}$
Work Step by Step
Let us consider that $I=\iiint_E 6xy dV$
Thus, $I= \int_{0}^1 \int_{0}^{\sqrt x} \int_{0}^{1+x+y} (6xy) dz dy dx$
Further, we have
$\int_{0}^1\int_{0}^{\sqrt x} |6xyz|_{0}^{1+x+y} dy dx=\int_{0}^1\int_{0}^{\sqrt x} (6xy+6x^2y+6xy^2) dy dx$
or, $=\int_0^1 [3xy^2+3x^2y^2+2xy^3]_{0}^{\sqrt x}dx$
or, $ \int^{0}_{1} 3x^2+3x^3+2x^{5/2} dx= [\dfrac{3x^3}{3}+\dfrac{3x^4}{4}+(2)(2/7)x^{7/2}]^{0}_{1}$
or, $=\dfrac{65}{28}$
